Does the answer change if the data source isn't a list, already in memory, but a stream? That is, will the sort end up pulling the entire stream into memory, when we only need k elements from the entire stream.
Interestingly, this is almost exactly the same as one of my standard interview questions, with the main difference being looking for the k elements closest to a target value, rather than the smallest. Not that it really changes the problem, but recognizing that is one of the things I'm looking for. On 4/12/07, apfelmus <[EMAIL PROTECTED]> wrote:
raghu vardhan <[EMAIL PROTECTED]>: > So, any algorithm that sorts is no good (think of n as huge, and k small). With lazy evaluation, it is! http://article.gmane.org/gmane.comp.lang.haskell.general/15010 [EMAIL PROTECTED] wrote: > The essence of all the answers that you've received is that it doesn't > matter if k is small, sorting is still the most elegant answer in Haskell. (It's not only most elegant, it can even be put to work.) > The reason is that in Haskell, a function which sort function will produce the > sorted list lazily. If you don't demand the while list, you don't perform > the whole sort. > > Some algorithms are better than others for minimising the amount of > work if not all of the list is demanded, but ideally, producing the > first k elements will take O(n log k + k) time. You mean O(k * log n + n) of course. Regards, apfelmus _______________________________________________ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
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